chauchitrung1996 wrote:Try to solve all problems in the Stewart's calculus book. If you solve them all quickly and understand them thoroughly, I believe calculus questions in mGRE would be a piece of cake (This is the only source I do calculus problems)
About Algebra and Topology, I do not know what to say. Hope someone would give a more comprehensive answer.
DDswife wrote:You should pay more attention to those. There are very interesting problems here or there
DDswife wrote:I solved a while ago this exercise, and I think that it’s GRE like
f(x) = x^2 + a |x| + 1
Find all possible a such that f(x) >= 0 for all x, x in R
a) (-oo,-2]
b) [-2,2]
c) [-2,oo)
d) [0,oo)
Enjoy
DDswife wrote:If you womder where I got this question from, it was in a Chinese website. It is in Chinese, of course. But I was able to follow the Math to understand what it meant. This is school level, according to what a Chinese frind of mine, who was the one that sent me the problems, told me. I believe that this is why they do good in the test. They are used to solve these weird questions since they are young,
Junaid456 wrote:I think the best way to solve this question is to do the relatively simple computations. There's a small catch in the question.
Assume that f(x) is greater than or equal to zero.
Case I: x is >= 0. In this case, f(x) = x^2 + ax + 1. Now, f(x) >= 0 iff f(x) has no roots, a repeated root, or distinct roots that are both negative. The first two conditions are equivalent to having the discriminant, D, less than equal to 0. D is a^2 - 4. So, D <= 0 iff a is between -2 and 2 (inclusive). If D is positive, we by the quadratic formula that the roots are -a + sqrt(a^2 - 4) < 0 (I have omitted the denominator) if and only if sqrt(a^2 - 4) < a. We must have a is greater than 0 over here, and by squaring, it is clear that the inequality is true (the other root is clearly negative). So, the last case is true if a is positive. So, a must be between -2 and infinity (inclusive). We must include all possibilities since f(x) >= 0 in 3 cases!
Case I: x is <= 0. In this case, f(x) = x^2 - ax + 1. Now, f(x) >= 0 iff f(x) has no roots, a repeated root, or distinct roots that are both negative. The first two conditions are equivalent to having the discriminant, D, less than equal to 0. D is a^2 - 4. Again, a is between -2 and 2 (inclusive). If D is positive, we by the quadratic formula that the roots are -a + sqrt(a^2 - 4) > 0 (I have omitted the denominator) if and only if a < sqrt(a^2 - 4). If a is positive, squaring, we see that this isn't true. If a is negative, this is always true, as sqrt(a^2 - 4) > 0. Considering the second root, -a + sqrt(a^2 - 4) > 0 (I have omitted the denominator) if and only if - a > sqrt(a^2 - 4) > 0. Again, a can't be positive. If a is negative, can square both sides, and see the inequality is true. So, a must be between - infinity and 2 (inclusive).
Both cases will be simultaneously true iff a is in [-2,2].
Please correct me if I am wrong.
It's easy to ignore the sub-case that double roots exists, but are in half in which we're not interested. This is what makes it to be an annoying mGRE like problem. On than that, we simply have to play with the cases, and use the basics of logic, which makes it a good problem to give in a pre-calculus course.
I haven't done it by playing with the answers by first trying to plug in -2, 2 (and 0). I guess the difficult thing in this approach is to figure out the behavior if a > 2 or a < -2, without doing any computations. I find it very hard, and that also takes a minute or two. It's better to write down the problem, but it's easy to screw it up if one wants to solve it in 2-ish minutes. I screwed it up before by not considering the sub-case in each case that I mentioned in the previous paragraph.
DDswife wrote:http://m.ishare.iask.sina.com.cn/f/63995575.html
I don’t understand what most of them mean, but you can guess and try to solve some of them
MMDE wrote:Junaid456 wrote:I think the best way to solve this question is to do the relatively simple computations. There's a small catch in the question.
Assume that f(x) is greater than or equal to zero.
Case I: x is >= 0. In this case, f(x) = x^2 + ax + 1. Now, f(x) >= 0 iff f(x) has no roots, a repeated root, or distinct roots that are both negative. The first two conditions are equivalent to having the discriminant, D, less than equal to 0. D is a^2 - 4. So, D <= 0 iff a is between -2 and 2 (inclusive). If D is positive, we by the quadratic formula that the roots are -a + sqrt(a^2 - 4) < 0 (I have omitted the denominator) if and only if sqrt(a^2 - 4) < a. We must have a is greater than 0 over here, and by squaring, it is clear that the inequality is true (the other root is clearly negative). So, the last case is true if a is positive. So, a must be between -2 and infinity (inclusive). We must include all possibilities since f(x) >= 0 in 3 cases!
Case I: x is <= 0. In this case, f(x) = x^2 - ax + 1. Now, f(x) >= 0 iff f(x) has no roots, a repeated root, or distinct roots that are both negative. The first two conditions are equivalent to having the discriminant, D, less than equal to 0. D is a^2 - 4. Again, a is between -2 and 2 (inclusive). If D is positive, we by the quadratic formula that the roots are -a + sqrt(a^2 - 4) > 0 (I have omitted the denominator) if and only if a < sqrt(a^2 - 4). If a is positive, squaring, we see that this isn't true. If a is negative, this is always true, as sqrt(a^2 - 4) > 0. Considering the second root, -a + sqrt(a^2 - 4) > 0 (I have omitted the denominator) if and only if - a > sqrt(a^2 - 4) > 0. Again, a can't be positive. If a is negative, can square both sides, and see the inequality is true. So, a must be between - infinity and 2 (inclusive).
Both cases will be simultaneously true iff a is in [-2,2].
Please correct me if I am wrong.
It's easy to ignore the sub-case that double roots exists, but are in half in which we're not interested. This is what makes it to be an annoying mGRE like problem. On than that, we simply have to play with the cases, and use the basics of logic, which makes it a good problem to give in a pre-calculus course.
I haven't done it by playing with the answers by first trying to plug in -2, 2 (and 0). I guess the difficult thing in this approach is to figure out the behavior if a > 2 or a < -2, without doing any computations. I find it very hard, and that also takes a minute or two. It's better to write down the problem, but it's easy to screw it up if one wants to solve it in 2-ish minutes. I screwed it up before by not considering the sub-case in each case that I mentioned in the previous paragraph.
If a > 0, f(x) will always be positive, here's a rough very informal proof:
x^2 > 0 for all x
a[x] > 0 for all x
Thus f(x) > 0 for all a >0, so there should be no upper bound on the interval. You can play around with the lower bound to find that it should be (I believe) [-2, oo).
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